$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}$

Question.

$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}$

solution:

Given $\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}$

Multiply and divide the given equation by $\sqrt{2}-\sqrt{1+\cos x}$ Then we get

$\Rightarrow$ $\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x} \times\left(\frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}\right)$

Now by splitting the limits we have

$\Rightarrow$ $\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x} \times\left(\frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}\right)=\lim _{x \rightarrow 0} \frac{2-(1+\cos x)}{\sin ^{2} x} \lim _{x \rightarrow 0}\left(\frac{1}{\sqrt{2}+\sqrt{1+\cos x}}\right)$

Now $\sin ^{2} x=1-\cos ^{2} x=(1-\cos x)(1+\cos x)$

Substituting this in above equation

$\Rightarrow x$ $\lim _{x \rightarrow 0} \frac{2-(1+\cos x)}{\sin ^{2} x} \lim _{x \rightarrow 0}\left(\frac{1}{\sqrt{2}+\sqrt{1+\cos x}}\right)=\frac{1}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{(1-\cos x)}{(1-\cos x)(1+\cos x)}$

Now by applying the limits we get

$\Rightarrow$ $\frac{1}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{(1-\cos x)}{(1-\cos x)(1+\cos x)}=\frac{1}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{1}{(1+\cos x)}=\frac{1}{2 \sqrt{2}} \cdot \frac{1}{2}$

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}=\frac{1}{4 \sqrt{2}}$

$\Rightarrow$$\frac{1}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{(1-\cos x)}{(1-\cos x)(1+\cos x)}=\frac{1}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{1}{(1+\cos x)}=\frac{1}{2 \sqrt{2}} \cdot \frac{1}{2}$

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}=\frac{1}{4 \sqrt{2}}$

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